Three identical springs (X, Y, Z)are arranged as shown.When a 4.0-kg mass is hung on X, the mass descends 3.0 cm.When a 6.0-kg mass is hung on Y, the mass descends:

Question

Quizplus · Accepted Answer

Since the springs are identical, the force constant k of each spring is the same. 
From Hooke's Law, we know that F = kx, where F is the force exerted by the spring, x is the displacement of the spring from its equilibrium position, and k is the force constant. 
For spring X, when a 4.0-kg mass is hung, we have F = mg = (4.0 kg)(9.81 m/s^2) = 39.24 N. 
Thus, kx = 39.24 N, where x = 0.03 m. 
Solving for k, we get k = 1308 N/m.

For spring Y, when a 6.0-kg mass is hung, we have F = mg = (6.0 kg)(9.81 m/s^2) = 58.86 N. 
Since the force constant k is the same for all three springs, we can use F = kx to find the displacement x of spring Y: 
58.86 N = (1308 N/m)x 
x = 0.045 m = 4.5 cm.

Therefore, the answer is E) 9.0 cm.

Three Identical Springs (X, Y, Z)are Arranged as Shown